The number of solutions of the equation sqrt{ | vedclass

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(B) Given equation: $\sqrt{3}\cos 2	heta + 8\cos 	heta + 3\sqrt{3} = 0$Using $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$\sqrt{3}(2\cos^2 	heta - 1

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$5$

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(B) Given equation: $\sqrt{3}\cos 2	heta + 8\cos 	heta + 3\sqrt{3} = 0$Using $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$\sqrt{3}(2\cos^2 	heta - 1) + 8\cos 	heta + 3\sqrt{3} = 0$$2\sqrt{3}\cos^2 	heta - \sqrt{3} + 8\cos 	heta + 3\sqrt{3} = 0$$2\sqrt{3}\cos^2 	heta + 8\cos 	heta + 2\sqrt{3} = 0$Dividing by $2$,we get $\sqrt{3}\cos^2 	heta + 4\cos 	heta + \sqrt{3} = 0$Factoring the quadratic: $(\sqrt{3}\cos 	heta + 1)(\cos 	heta + \sqrt{3}) = 0$This gives $\cos 	heta = -\frac{1}{\sqrt{3}}$ or $\cos 	heta = -\sqrt{3}$.Since $-1 \le \cos 	heta \le 1$,the value $\cos 	heta = -\sqrt{3}$ is rejected.For $\cos 	heta = -\frac{1}{\sqrt{3}}$,we look for solutions in the interval $[-3\pi, 2\pi]$.In the interval $[0, 2\pi]$,there are $2$ solutions.In the interval $[-2\pi, 0]$,there are $2$ solutions.In the interval $[-3\pi, -2\pi]$,there is $1$ solution.Total number of solutions $= 2 + 2 + 1 = 5$.

The number of solutions of the equation $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0$ for $\theta \in [-3\pi, 2\pi]$ is:

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